Identify `Z` in the following series
`CH_(2)=CH_(2) overset(HBr)rarr X overset(Hydrolysis)rarr Y overset(Na_(2) CO_(3))underset(I_(2)excess)rarrZ`

Options

A. `C_2 H_5 I`
B. `C_2 H_5 OH`
C. `CHI_3`
D. `CH_3 CHO`

Explanation

To identify the compound `Z` in the given series, we will follow the steps outlined in the question and analyze each reaction.

### Step 1: Identify the starting compound
The starting compound is ethene, represented as \( CH_2=CH_2 \).

### Step 2: Reaction with HBr
When ethene reacts with HBr, the addition of HBr across the double bond occurs. Since ethene is symmetrical, we can add H to either carbon atom and Br to the other. This results in the formation of bromoethane (X).

**Reaction:**
\[ CH_2=CH_2 + HBr \rightarrow CH_3CH_2Br \]
- **X = Bromoethane (C2H5Br)**

### Step 3: Hydrolysis of X
Next, we perform hydrolysis on bromoethane (X). Hydrolysis of an alkyl halide results in the formation of an alcohol. In this case, bromoethane will be hydrolyzed to form ethanol (Y).

**Reaction:**
\[ CH_3CH_2Br + H_2O \rightarrow CH_3CH_2OH + HBr \]
- **Y = Ethanol (C2H5OH)**

### Step 4: Reaction of Y with Na2CO3 and I2
Now, we take ethanol (Y) and react it with sodium carbonate (Na2CO3) and excess iodine (I2). This reaction is known as the iodoform reaction. Ethanol, when treated with iodine in the presence of a base, will produce iodoform (Z).

**Reaction:**
\[ CH_3CH_2OH + I_2 + Na_2CO_3 \rightarrow CHI_3 + Na_2CO_3 + H_2O \]
- **Z = Iodoform (CHI3)**

### Conclusion
Thus, the compound `Z` in the given series is **Iodoform (CHI3)**.

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